Question

The equations to the circles which touch the lines 3x – 4y + 1 = 0, 4x + 3y – 7 = 0 and pass through (2, 3) are

• A. $x^2$ + $y^2$ - 4x - 16y + 43 = 0, 5$x^2$ + 5$y^2$ - 12x - 24y + 31 = 0
• B. $x^2$ + $y^2$ + 4x + 16y - 43 = 0, 5$x^2$ + 5$y^2$ - 12x - 24y + 31 = 0
• C. $x^2$ + $y^2$ - 4x - 16y + 43 = 0, 5$x^2$ + 5$y^2$ + 12x + 24y + 31 = 0
• D. $x^2$ + $y^2$ + 4x + 16y - 43 = 0, 5$x^2$ + 5$y^2$ + 12x + 24y + 31 = 0

Answer 1

The correct option is A $x^2$ + $y^2$ - 4x - 16y + 43 = 0, 5$x^2$ + 5$y^2$ - 12x - 24y + 31 = 0

Since the circle touches the given lines which are perpendicular, centre of the circle lies on the bisectors of angels between the given lines.
The bisectors of angles between the given lines are $\frac{3x-4y+1}{5}$ $\pm$ $\frac{4x+3y-7}{5}$ = 0
$\Rightarrow$ (3x - 4y + 1) + (4x + 3y - 7) = 0 or (3x - 4y + 1) - (4x + 3y - 7) = 0
$\Rightarrow$ 7x - y - 6 = 0 or - x - 7y + 8 = 0 $\Rightarrow$ y = 7x - 6 or x = 8 - 7y
If the centre lies on x = 8 - 7y, then the centre is of the form (8 - 7k, k)
$\therefore$ $\left|\frac{3(8-7k)-4k+1}{5}\right|$ = $\sqrt{{(8-7k-2)}^2+{(k-3)}^2}$ $\Rightarrow (24-25k+1{)}^2=25(6-7k{)}^2+25(k-3{)}^2$
$\Rightarrow {25}^2(1-5{)}^2=25[36+49k^2-84k+k^2+9-6k]$ $\Rightarrow 25(1+k^2-2k)=50k^2-90k+45$
$\Rightarrow 25k^2$ - 40k + 20 = 0 $\Rightarrow 5k^2$ - 8k + 4 = 0 $\Rightarrow$ k is imaginary
$\therefore$ The centre of the circle lies on the line y = 7x - 6 and hence it is of the form (k, 7k - 6)
$\therefore$ $\left|\frac{3k-k(7k-6)+1}{5}\right|$ = $\sqrt{{(k-2)}^2+{(7k-6-3)}^2}$ $\Rightarrow$ (3k - 28k + 24 - 1)${}^2$ = 25 [(k - 2)${}^2$ + (7k - 9)${}^2$]
$\Rightarrow$ (25 - 25k)${}^2$ = 25(k${}^2$ + 4 - 4k + 49k2 + 81 - 126k) $\Rightarrow$ 25(1 + k${}^2$ - 2k) = 50k${}^2$ - 130k + 85
$\Rightarrow$ 25k${}^2$ - 80k + 60 = 0 $\Rightarrow$ 5k${}^2$ - 16k + 12 = 0 $\Rightarrow$ (k - 2) (5k - 6) = 0 $\Rightarrow$ k = 2 or k = $\frac{6}{5}$
If k = 2, then centre = (2, 8), radius = 5
Equation of the circle is $(x-2{)}^2+(y-8{)}^2$ = 25 $\Rightarrow$ $x^2$ + $y^2$ - 4x - 16y + 43 = 0
If k = $\frac{6}{5}$, then centre is $(\frac{6}{5},\frac{12}{5})$, radius = $\sqrt{{\left(\frac{4}{5}\right)}^2+{\left(\frac{3}{5}\right)}^2}$ = 1
Equation of the circle is ${(x-\frac{6}{5})}^2$ + ${(y-\frac{12}{5})}^2=1\Rightarrow (5x-6{)}^2+(5y-12{)}^2$ = 25
$\Rightarrow$ 25$x^2$ + 25$y^2$ - 60x - 120y + 155 = 0 $\Rightarrow$ 5$x^2$ + 5$y^2$ - 12x - 24y + 31 = 0
$\therefore$ Required circles are $x^2$ + $y^2$ - 4x - 16y + 43 = 0 and 5$x^2$ + 5$y^2$ - 12x - 24y + 31 = 0